Ответ(ы) на вопрос:
Гость1. b. 2cos(x/2)-1=0, 2cos(x/20=1. cos(x/2)=1/2
x/2=+-arccos(1/2)+2πn, n∈Z
x/2=+-π/3+2πn, n∈Z |*2
x=+-2π/3+4πn, n∈Z
г. √(1-cosx)=sinx
ОДЗ: 1-cosx≥0, -cosx≥-1. cosx≤0. x∈[π/2+2πn;3π/2+2πn], n∈Z
sinx>0. x∈(2πn;π+2πn), n∈Z
=> x∈[π/2+2πn;π+2πn), n∈Z
(√(1-cosx))²=(sinx)², 1-cosx=sin²x. 1-cosx=1-cos²x. cos²x-cosx=0
cosx*(coxs-1)=0
cosx=0 или cosx-1=0
1. cosx=0. x=π/2+πn, n∈Z
2. cosx=1. x=2πn ,n∈Z
3. {cosx+cosxy=1 {cosx+cosy=1 {cos(2π-y)+cosy=1 {cosy+cosy=1
x+y=2π x=2π-y x=2π-y x=2π-y
{2cosy=1 {cosy=1/2
x=2π-y x=2π-y
cosy=1/2. y=+-arccos(1/2)+2πn, n∈Z. y=+-π/3+2πn, n∈Z
{y=-π/3+2πn, n∈Z {x₁=7π/3-2πn, n∈Z
x=2π-(-π/3+2πn), n∈Z y₁=-π/2+2πn, n∈Z
{y=π/3+2πn, n∈Z {x₂=5π/3-2πn, n∈Z
x=2π-(π/3+2πn), n∈Z y₂=π/3+2πn, n∈Z
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